Problem: You have found the following ages (in years) of 6 tigers. Those tigers were randomly selected from the 23 tigers at your local zoo: $ 20,\enspace 15,\enspace 14,\enspace 5,\enspace 3,\enspace 23$ Based on your sample, what is the average age of the tigers? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 23 tigers, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\overline{x}} = \dfrac{20 + 15 + 14 + 5 + 3 + 23}{{6}} = {13.3\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {44.89} + {2.89} + {0.49} + {68.89} + {106.09} + {94.09}} {{6 - 1}} $ {s^2} = \dfrac{{317.34}}{{5}} = {63.47\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{63.47\text{ years}^2}} = {8\text{ years}} $ We can estimate that the average tiger at the zoo is 13.3 years old. There is also a standard deviation of 8 years.